Polar Equation of an Ellipse

Date: 12/16/95 at 22:30:56
From: Anonymous
Subject: Geometry
Find the polar equation of the ellipse with eccentricity 3/4,
one focus at the pole, and the corresponding directrix
perpendicular to the polar axis, through the point with
polar coordinates (8,pi).

Date: 5/30/96 at 14:59:10
From: Doctor Charles
Subject: Re: Geometry
The distance of a point on the ellipse from the directrix is (x-8) and
its distance from the origin is sqrt(x^2+y^2). So by the definition of
the ellipse:
(x-8)^2=e^2*(x^2+y^2) e is the eccentricity
So in polar coordinates:
(r cos(t) - 8)^2 = r^2 * e^2
r^2 (cos^2 (t) - e^2) - r * 16 cos(t) + 64 = 0
This quadratic can be solved for r in terms of t. (apply the formula)
-Doctor Charles, The Math Forum

Date: 5/30/96 at 15:0:12
From: Doctor Anthony
Subject: Re: Geometry
Let S be the focus of the conic and KL its directrix. SZ is the
perpendicular from S on KL, and SZ is 180 degrees from the initial
line, with S as the pole of coordinates. P=(r,theta) is any point on the
conic and ST is the semi-latus rectum of length L. The eccentricity of
the conic is e, and TR is the perpendicular from T to KL. PN is the
ordinate of P, and PM is the perpendicular from P to KL.
From the definition of a conic, ST = e.TR, that is, L = e.TR
so TR = L/e
From the diagram SZ = TR = L/e
Now SN = r.cos(theta) and SP = r = e.PM
Therefore PM = NZ = r/e
Also SZ = NZ - NS
Therefore L/e = r/e - r.cos(theta)
So the polar equation is
L/r = 1 - e.cos(theta)
To find the polar equation of the directrix, using same diagram let
X = (r,theta) be a point on the directrix KL
SZ = SX.cos(pi-theta) = -SX.cos(theta)
= -r.cos(theta)
But SZ = L/e, therefore L/e = -r.cos(theta) is the equation of the
directrix.
We are given that e = 3/4 and that when theta=pi, r=8
So L/(3/4) = -8.cos(pi). From this L = (3/4)(8) = 6
The equation of our ellipse is:
6/r = 1 - (3/4).cos(theta)
-Doctor Anthony, The Math Forum